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Chapter 1: Problem 32

Suppose all \(n\) men at a party throw their hats in the center of the room. Each man then randomly selects a hat. Show that the probability that none of the \(n\) men selects his own hat is $$ \frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-+\cdots \frac{(-1)^{n}}{n !} $$ Note that as \(n \rightarrow \infty\) this converges to \(e^{-1}\). Is this surprising?

Short Answer

Expert verified
The probability that none of the n men selects his own hat is given by the alternating sum of the reciprocals of factorials: $$ P(\text{no man selects his own hat}) = \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots \frac{(-1)^n}{n!} $$ As n approaches infinity, this probability converges to \(e^{-1}\).

Step by step solution

01

Derangement Concept

A derangement, in the context of permutations, is a permutation with no fixed points, i.e., there is no element that remains in its original position after permuting. In this exercise, we will be finding the probability that none of the n men selects his own hat, which translates to finding the derangement of n elements.
02

Counting Total Permutations and Derangements

There are n! (factorial n) possible ways to arrange the hats for n men. Let D_n denote the number of derangements for a given n. There are various ways to compute D_n, but we will use the principle of inclusion and exclusion.
03

Inclusion-Exclusion Principle

According to the principle of inclusion and exclusion, the total number of derangements can be calculated by alternating sum of the factorials: $$ D_n = n! - \binom{n}{1}(n-1)! + \binom{n}{2}(n-2)! - \binom{n}{3}(n-3)! + \cdots + (-1)^n\binom{n}{n}(0)! $$
04

Simplifying the Derangement Formula

To find the probability that none of the n men selects his own hat, we will divide the number of derangements by the total number of permutations (n!). $$ P(\text{no man selects his own hat}) = \frac{D_n}{n!} $$ Plugging in the derangement formula from step 3, we have: $$ P(\text{no man selects his own hat}) = \frac{n! - \binom{n}{1}(n-1)! + \binom{n}{2}(n-2)! - \binom{n}{3}(n-3)! + \cdots + (-1)^n\binom{n}{n}(0)!}{n!} $$
05

Cancelling n! from the Derangement Formula

We can simplify this further by cancelling n! from both the numerator and denominator: $$ P(\text{no man selects his own hat}) = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!} $$ So the probability that none of the n men selects his own hat is: $$ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots \frac{(-1)^n}{n!} $$ As the exercise notes, this probability converges to \(e^{-1}\) as n approaches infinity.

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Most popular questions from this chapter

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